A 1-kW radio transmitter operates at a frequency of 880 Hz. How many photons per second does it emit?

Options

A. `1.71 xx 10^(21)`
B. `1.71 xx 10^(30)`
C. `6.02 xx 10^(23)`
D. `1.71 xx 10^(33)`

Explanation

To find the number of photons emitted per second by a 1-kW radio transmitter operating at a frequency of 880 Hz, we can follow these steps:

### Step 1: Understand the relationship between energy, number of photons, and frequency
The energy emitted per second (power) by the transmitter can be expressed in terms of the number of photons emitted and the energy of each photon. The formula is:
\[ E = N \cdot h \cdot \nu \]
where:
- \( E \) is the total energy (in joules per second, which is power),
- \( N \) is the number of photons emitted per second,
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{J s} \)),
- \( \nu \) is the frequency (in Hz).

### Step 2: Convert power to joules per second
The power of the transmitter is given as 1 kW, which is equivalent to:
\[ E = 1 \, \text{kW} = 1000 \, \text{W} = 1000 \, \text{J/s} \]

### Step 3: Substitute known values into the equation
Now, we can rearrange the formula to solve for \( N \):
\[ N = \frac{E}{h \cdot \nu} \]
Substituting the known values:
- \( E = 1000 \, \text{J/s} \)
- \( h = 6.626 \times 10^{-34} \, \text{J s} \)
- \( \nu = 880 \, \text{Hz} \)

### Step 4: Calculate the number of photons
Now, plug in the values:
\[ N = \frac{1000}{(6.626 \times 10^{-34}) \cdot (880)} \]

Calculating the denominator:
\[ h \cdot \nu = 6.626 \times 10^{-34} \cdot 880 = 5.82888 \times 10^{-31} \, \text{J} \]

Now calculate \( N \):
\[ N = \frac{1000}{5.82888 \times 10^{-31}} \]

Calculating this gives:
\[ N \approx 1.71 \times 10^{33} \]

### Final Answer
Thus, the number of photons emitted per second by the transmitter is approximately:
\[ N \approx 1.71 \times 10^{33} \]